Question 102

What is the simplified value of $$(3 + 1)(3^{2} + 1)(3^{4} + 1)(3^{8} + 1)(3^{16} + 1)$$?

Solution

Given $$(3 + 1)(3^{2} + 1)(3^{4} + 1)(3^{8} + 1)(3^{16} + 1)$$

Multiply and divide by (3 - 1)

$$\frac{(3 - 1)(3 + 1)(3^{2} + 1)(3^{4} + 1)(3^{8} + 1)(3^{16} + 1)}{3 - 1}$$

$$\frac{(3^{2} -1)(3^{2} + 1)(3^{4} + 1)(3^{8} + 1)(3^{16} + 1)}{2}$$

$$\frac{(3^{8} - 1)(3^{8} + 1)(3^{16} + 1)}{2}$$

$$\frac{(3^{16} - 1)(3^{16} + 1)}{2}$$

$$\frac{3^{32} - 1}{2}$$

Hence, option A is the correct answer.

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